Redox and Electroplating

How Are Redox Reactions Different?

Redox is the term used to label reactions in which the
acceptance of an electron (reduction) by a material is matched
with the donation of an electron (oxidation). A large number of
the reactions already mentioned in the Reactions chapter are
redox reactions.

Synthesis reactions are also redox
reactions if there is an exchange of electrons to make an ionic
bond. If chlorine gas is added to sodium metal to make sodium
chloride, the sodium has donated an electron and the chlorine has
accepted an electron to become a chloride ion or an attached
chlorine.

If a compound divides into elements in a decomposition, a decomposition reaction could be a redox
reaction. The electrolysis of water is a redox reaction. With a
direct electric current through it, water can be separated into
oxygen and hydrogen. H2O
===>
H2 + O2 The
oxygen and hydrogen in the water are attached by a covalent bond
that breaks to make the element oxygen and the element hydrogen.
Learning more about the conditions for redox reactions will show
that the electrolysis of water is a redox reaction.

A single replacement reaction
is always a redox reaction because it involves an element that
becomes incorporated into a compound and an element in the
compound being released as a free element.

A double replacement reaction
usually is not a redox reaction.

Oxidation States

Before we go any further into redox, we must understand
oxidation states. The idea of oxidation state began with whether
or not a metal was attached to an oxygen. Unattached (free) atoms
have an oxidation state of zero. Since oxygen almost
always takes in two electrons when it is not a free element, the
combined form of oxygen (oxide) has an oxidation state of
minus two. The exception to a combined oxygen taking two
electrons is the peroxide configuration. Peroxide can be
represented by -O-O- where the each dash is a covalent bond and
each ëOí is an oxygen atom. Peroxide can be written
as a symbol, (O2)2-. The over-simplified way of showing
this is that each oxygen atom has a negative one oxidation state,
but that is not really so because the peroxides do not come in
individual oxygen atoms. Peroxides are not as stable as oxides,
and there are very many fewer peroxides in nature than oxides.
H2O2 is hydrogen
peroxide.

Hydrogen in compound always has an oxidation state of plus one,
except as a hydride. A hydride is a compound of a metal and
hydrogen. The hydrogen atoms in a hydride have the oxidation state of -1. Hydrides react with water, so there are no hydrides
found in nature. The formula XH or XH2 or
XH3 or even XH4 where X
is a metal is the general chemical formula for hydride.

The rules for oxidation state are in some ways arbitrary and
unnatural, but here they are:

1. Any free (unattached) element with no charge has the
oxidation state of zero. Diatomic gases such as O2 and H2 are
also in this category.

2. All compounds have a net oxidation state of zero. The
oxidation state of all of the atoms add up to zero.

3. Any ion has the oxidation state that is the charge of
that ion. Polyatomic ions (radicals) have an oxidation state for
the whole ion that is the charge on that ion. The ions of
elements in Group I, II, and VII (halogens) and some other elements only
have one likely oxidation state.

4. Oxygen in compound has an oxidation state of minus two,
except for oxygen as peroxide, which is minus one.

5. Hydrogen in compound has an oxidation state of plus one,
except for hydrogen as hydride, which is minus one.

6. In radicals or small covalent molecules, the
element with the greatest electronegativity has its natural ion
charge as its oxidation state.

KNOW THIS

Now would be a good time to try the oxidation state problems
beginning the practice page at the end of this chapter.
Problems 1-30 are good examples for practice of assigning
oxidation states.

Is It a Redox Reaction?

A redox reaction will have at least one type of atom releasing
electrons and another type of atom accepting electrons. How
can you most easily tell if a reaction is redox? Label every atom
on both the reactant and product side of the equation with
its oxidation number. If there is a change in oxidation number
from one side of the equation to the other of the same species
of atom, it is a redox reaction. Each complete equation must have
at least one atom species losing electrons and at least
one atom species gaining electrons. The loss and gain of
electrons will be reflected in the changes of oxidation number.

Letís take the following equation: K2(Cr2O7) + KOH
===>
2 K2(CrO4) + H2O Is it a redox equation or not?
Potassium dichromate and potassium hydroxide make potassium
chromate and water. Some of the atoms are easy. All of the
oxygens in compound have an oxidation state of minus two. All of
the hydrogens have an oxidation state of plus one. Potassium is a
group one element, so it should have an oxidation state of plus
one in the compounds. That seems to make sense because dichromate
and chromate ions have a charge of minus two and there are two
potassium atoms in each compound. Hydroxide ion has a charge of
minus one and it has one potassium. But what about the chromium
atoms? We can do a little primitive math on the material either
from the starting point of the compound or the ion to find the
oxidation state of chromium in that compound. The entire compound
must have a net oxidation state of zero, so the oxidation numbers
of two potassiums one chromium and four oxygens must equal to
zero. 2 K + Cr + 4 O = 0 We know the oxidation state of
everything else but the chromium. 2(+1) + Cr + 4 (-2) = 0 and Cr
= +6. Or we could do it from the point of view of the
chromate ion. Cr + 4 O = -2 The oxygens are minus two each. Cr +
4 (-2) = -2 Either way Cr = +6. Now the dichromate; 2 K + 2 Cr +
7 O = 0 and 2 (+1) + 2 Cr + 7 (-2) = 0. Then 2 Cr = +12 and Cr =
+6. You can do the math for the dichromate ion to see for
yourself that the chromium does not change from one side of this
equation to the other. As suspicious-appearing as the equation
might have seemed to you, it is not a redox reaction.

Consider copper metal in silver nitrate solution becomes
silver metal and copper II nitrate. The oxygens do not change.
Oxygen in compound is negative two on both sides. The nitrogen
cannot change. It does not move out of the nitrate ion
where it has an oxidation state of plus five. (Is that right?)
The other two have to change because they both are elements
with a zero oxidation state on one side and in compound on the
other. Silver goes from plus one to zero and Copper goes
from zero to plus two. AgNO3 + Cu
===>
Cu(NO3)2 + Ag (not a balanced reaction)

Think of this on a number line. The copper is oxidized because
its oxidation number goes up from zero to plus two. The silver is
reduced because its oxidation number reduces from plus one to
zero.

Half Reactions

Half reactions are either an oxidation or a reduction. Only
the species of atom that is involved in a change is in a half
reaction. In the above reaction, silver goes from plus one to
zero oxidation state, but to account for everything, the
electrons must be placed into the half reaction. e- + Ag+
===>
Ag0 (Reduction) Notice that the half
reaction must be balanced in charge also and that the only way to
balance it is to add electrons to the more positive side. The
other half reaction is that of copper. Cu0
===>
Cu+2 + 2 e- (Oxidation)

This time the material is oxidized and the electrons must
appear on the product side. We must double the silver half
reaction to cancel out the electrons from right to left. The two
half reactions can be added together to make one reaction, thus.

In the complete reaction the number of electrons lost must equal the number of electrons gained. The number of electrons used in the reduction half reaction must equal the number of electrons produced in the oxidation half reaction. The entire half reactions must be multiplied by numbers that will equalize the numbers of electrons, and the final complete balanced chemical reaction must show these number relationships.

One of the important bits of information from adding the half
reactions in this case is that the entire chemical equation will have to have
two silver atoms for every copper atom in the reaction for the
reaction to balance electrically. This type of information from
the half reactions is sometimes the easiest or only way to
balance a chemical equation. The redox balancing problems
beginning with number 31 at the end of the chapter are good help
for your further understanding.

From doing this math on a number of materials, you will find
that it is possible to get some strange-looking oxidation
states, to include some fractional ones. The oxidation state math
works on fractional oxidation states also, even though fractional
charges are not possible.

Reduction or Oxidation?

A reduction of a material is the gain of electrons. An
oxidation of a material is the loss of electrons. This system
comes from the observation that materials combine with oxygen in
varying amounts. For instance, an iron bar oxidizes (combines
with oxygen) to become rust. We say that the iron has oxidized.
The iron has gone from an oxidation state of zero to (usually)
either iron II or iron III. This may be difficult to remember.
The easier way to tell if a half reaction is a reduction or
oxidation is to plot the changing ion into the number line. If
the oxidation state of the ion goes up the number line, it is an
oxidation. If it goes down the number line, it is a reduction.
Based on the KIS principle (Keep It Simple), remember only one
rule for this.

Someone, in a fit of perversity, decided that we needed more
description for the process. A material that becomes oxidized is
a reducing agent, and a material that becomes reduced is an
oxidizing agent.

COMING ATTRACTIONS

ELECTROLYSIS
Water, aluminum, copper

ELECTROPLATING
chromium, gold,

ELECTRIC CELLS
carbon, zinc, etc electrical potential and voltages

Redox Problems

For each element in the
following materials list the number of the rule you use to assign
oxidation state of that element and list the oxidation state you
have found it to be.

.

Summary of
Oxidation State Rules

1. Free element O.S. = 0

2.
Compound total O.S. = 0

3. Ion O.S. =
charge

4. Oxygen O.S. = -2

5.
Hydrogen O.S. = +1

6. Electronegativity
rules

.

MATERIAL

RULES

OXIDATION STATES

1. NaCl

Na 3; Cl 3

Na = +1 , Cl = -1

2. KMnO4

K 3; O 4; Mn 3,4 or
2,3,4

K = +1, O = -2, Mn = +7

3. diamond

C 1

C = 0

4. CO2

O 4; C 4,2

C
= +4, O = -2

5. CO

O 4; C 4,2

C = +2, O = -2

6. KCN

K 3; N 6; C 3,6, 2

K = +1, C =
+2, N=-3

7. Na4Fe(CN)6

Na 3; N 6; C
6,3; Fe 3,6,2

Na= +1, N= -3, C= +2, Fe= +2

8. Fe2O3

O 4; Fe 4,2

O = -2, Fe = +3

9. Fe3O4

O 4; Fe 4,2

O = -2, Fe = +8/3

10. (ClO4)-

O 4; Cl
4,3

O = -2, Cl = +7

11. (ClO3)-

O 4; Cl 4,3

O = -2, Cl = +5

12. (ClO2)-

O 4; Cl 4,3

O = -2, Cl = +3

13. (ClO)-

O 4; Cl 4,3

O = -2, Cl = +1

14. Cl-

Cl 3

Cl = -1

15. Cl2

Cl 1

Cl = 0

16. P2O5

O 4; P 4,2

O = -2, P = +5

17. P4O6

O 4; P 4,2

O = -2, P = +3

18. H3PO4

H 5; O 4; P 5,4,2 or 4,3

H = +1, O = -2,
P = +5

19. Mg3N2

Mg 3; N 3,2

Mg = +2, N =
-3

20. MgH2

Mg 3; H 3,2

Mg = +2, H = -1 (hydride!)

21. NH3

H 5; N 5,2

H
= +1, N = -3

22. N2H4

H 5; N 5,2

H = +1, N = -2

23. (NH4)+

H 5; N 5,2

H = +1, N = -3

24. N2

N 1

N = 0

25. (NO3)-

O 4; N
4,3

O = -2, N = +5

26. (NO2)-

O 4; N 4,3

O = -2, N = +3

27. NO2

O 4; N 4,2

O = -2, N = +4

28. NO

O 4; N 4,2

O = -2, N = +2

29. N2O

O 4; N 4,2

O
= -2, N = +1

30. Na2O2

Na 3; O 2

Na = +1, O = -1 (peroxide!)

Redox Problems Part 2

B. For the each word reaction,
write the chemical equation without balancing it, write the
oxidation state of each element above that element, and write the
two half reactions, labeling which is oxidation and which is
reduction. You can check your work by balancing the complete
reaction using the numbers from the half reaction addition. If
you have a problem with an example, check first with the
completed balanced equation in the answer section.

31. Hydrogen gas burns in oxygen to make water.

32. Mercuric oxide, a red powder, is put into a test tube and warmed. Liquid mercury forms on the sides and in the bottom of the tube and oxygen gas escapes from the test tube.

33. Potassium chlorate is heated in a test tube. Oxygen gas is made and potassium chloride is left in the bottom of the tube.